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Parallel and Perpendicular Lines

What you'll learn

  • How to read the gradient from a straight-line equation.
  • How to write equations of parallel lines.
  • How to write equations of perpendicular lines.
  • How to use two points, or an unknown value like kkk, to find a line’s gradient.

1. Start with y=mx+cy=mx+cy=mx+c

Most questions in this topic are about comparing gradients. So your first job is often to get the equation into the form y=mx+cy=mx+cy=mx+c.

Definition

Gradient and y-intercept

  • The gradient tells you the steepness and direction of a straight line. In y=mx+cy=mx+cy=mx+c, the gradient is mmm.

A line in the form y = mx + c, showing the gradient as rise divided by run and the y-intercept c.

  • The y-intercept is where the line crosses the y-axis. In y=mx+cy=mx+cy=mx+c, the y-intercept is ccc.

“Make yyy the subject” means rearrange the equation so that yyy is on its own.

Example

Finding the gradient

  1. Start with the equation 6y=3x−126y=3x-126y=3x−12.

  2. Divide every term by 6 to make yyy the subject: y=12x−2y=\frac{1}{2}x-2y=21​x−2.

  3. Compare it with y=mx+cy=mx+cy=mx+c.

  4. The gradient is m=12m=\frac{1}{2}m=21​ and the y-intercept is c=−2c=-2c=−2.

Common Mistake

Reading the wrong coefficient

In 2x+5y=102x+5y=102x+5y=10, the coefficient of xxx is not the gradient yet. Rearrange first until yyy is on its own.

2. Parallel lines

Definition

Parallel lines

Parallel lines are straight lines that never meet. In y=mx+cy=mx+cy=mx+c form, parallel lines have the same gradient.

Parallel lines have the same steepness but different y-intercepts.

The y-intercept can change, but the gradient must stay the same.

Example

Parallel line through the y-axis

  1. The line y=−3x+8y=-3x+8y=−3x+8 has gradient m=−3m=-3m=−3.

  2. A parallel line has the same gradient, so start with y=−3x+cy=-3x+cy=−3x+c.

  3. Passing through (0, 5) means the y-intercept is 5, so c=5c=5c=5.

  4. The equation is y=−3x+5y=-3x+5y=−3x+5.

Tip

When it says “a line”

If no point is given, there are many correct answers. For example, a line parallel to y=3x+1y=3x+1y=3x+1 could be y=3x−4y=3x-4y=3x−4.

3. Perpendicular lines

Definition

Perpendicular lines

Perpendicular lines meet at a right angle, 90°. For usual straight lines in y=mx+cy=mx+cy=mx+c form, their gradients are negative reciprocals, meaning you flip the fraction and change the sign. Their gradients satisfy m1m2=−1m_1m_2=-1m1​m2​=−1.

Perpendicular lines meet at 90°, and their gradients are negative reciprocals.

For example:

  • m=4m=4m=4 becomes m=−14m=-\frac{1}{4}m=−41​.
  • m=−23m=-\frac{2}{3}m=−32​ becomes m=32m=\frac{3}{2}m=23​.
Example

Perpendicular line through the y-axis

  1. The line y=14x−2y=\frac{1}{4}x-2y=41​x−2 has gradient m=14m=\frac{1}{4}m=41​.

  2. A perpendicular line has gradient m=−4m=-4m=−4.

  3. Passing through (0, -6) means the y-intercept is -6, so c=−6c=-6c=−6.

  4. The equation is y=−4x−6y=-4x-6y=−4x−6.

Common Mistake

Only doing half the reciprocal rule

For perpendicular gradients, you must do both parts: flip the fraction and change the sign.

Common Mistake

Horizontal and vertical lines

The negative-reciprocal rule assumes both lines can be written as y=mx+cy=mx+cy=mx+c. A horizontal line like y=4y=4y=4 is perpendicular to a vertical line like x=−1x=-1x=−1.

A horizontal line and a vertical line are perpendicular, even though the vertical line is not in y = mx + c form.

4. If the equation is not already in y=mx+cy=mx+cy=mx+c

Sometimes the given line is written in a different form, such as 3x+6y=123x+6y=123x+6y=12. Always rearrange first, then use the parallel or perpendicular rule.

Example

Rearrange first, then use the rule

  1. Rearrange 3x+6y=123x+6y=123x+6y=12 by subtracting 3x3x3x: 6y=−3x+126y=-3x+126y=−3x+12.

  2. Divide by 6: y=−12x+2y=-\frac{1}{2}x+2y=−21​x+2.

  3. The given line has gradient m=−12m=-\frac{1}{2}m=−21​.

  4. A perpendicular line has gradient m=2m=2m=2.

  5. Passing through (0, 7) gives c=7c=7c=7.

  6. The equation is y=2x+7y=2x+7y=2x+7.

5. Using two points to find a gradient

If a line is described using two points, calculate its gradient first.

Definition

Gradient between two points

For points (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​), the gradient is m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}m=x2​−x1​y2​−y1​​: change in yyy divided by change in xxx.

Gradient between two points is the vertical change divided by the horizontal change.

Once you know the gradient, use y=mx+cy=mx+cy=mx+c. To substitute a point means to replace xxx and yyy with that point’s coordinates.

Example

Parallel line from two points

  1. Line A passes through (1, 2) and (5, 14), so its gradient is m=14−25−1=3m=\frac{14-2}{5-1}=3m=5−114−2​=3.

Line A’s two points show a rise of 12 and a run of 4, giving gradient 3; the required parallel line has the same slope through (2, 3).

  1. A parallel line also has gradient m=3m=3m=3.

  2. Start with y=3x+cy=3x+cy=3x+c.

  3. Use the point (2, 3): 3=3(2)+c3=3(2)+c3=3(2)+c.

  4. Solve to get c=−3c=-3c=−3.

  5. The equation is y=3x−3y=3x-3y=3x−3.

Example

Perpendicular line from two points

  1. Line A passes through (0, 4) and (6, 7), so its gradient is m=7−46−0=12m=\frac{7-4}{6-0}=\frac{1}{2}m=6−07−4​=21​.

The line through (1, 5) is perpendicular to Line A, so it crosses at a right angle with the negative reciprocal gradient.

  1. A perpendicular line has gradient m=−2m=-2m=−2.

  2. Start with y=−2x+cy=-2x+cy=−2x+c.

  3. Use the point (1, 5): 5=−2(1)+c5=-2(1)+c5=−2(1)+c.

  4. Solve to get c=7c=7c=7.

  5. The equation is y=−2x+7y=-2x+7y=−2x+7.

Common Mistake

Gradient fraction upside down

Gradient is change in yyy over change in xxx. Keep the order of the two points consistent in the numerator and denominator.

6. Showing lines are parallel or perpendicular

When a question says “show that”, you need evidence. Calculate both gradients and then make a clear statement.

Example

Showing two lines are parallel

  1. Line A passes through (1, 2) and (3, 8), so mA=8−23−1=3m_A=\frac{8-2}{3-1}=3mA​=3−18−2​=3.

  2. Line B passes through (-2, 1) and (0, 7), so mB=7−10−(−2)=3m_B=\frac{7-1}{0-(-2)}=3mB​=0−(−2)7−1​=3.

  3. The gradients are equal, so the two lines are parallel.

For perpendicular lines, show that the gradients multiply to -1.

7. Finding an unknown value like kkk

If a coordinate contains kkk, write the gradient using kkk, then set it equal to the gradient you need.

Example

Finding k using perpendicular gradients

  1. Line A passes through (-1, 2) and (3, 4), so mA=4−23−(−1)=12m_A=\frac{4-2}{3-(-1)}=\frac{1}{2}mA​=3−(−1)4−2​=21​.

  2. A perpendicular line must have gradient m=−2m=-2m=−2.

  3. Line B passes through (2, 5) and (kkk, -1), so its gradient is mB=−1−5k−2=−6k−2m_B=\frac{-1-5}{k-2}=\frac{-6}{k-2}mB​=k−2−1−5​=k−2−6​.

Line B uses an unknown x-coordinate k, so its horizontal change is k - 2 while its vertical change is -6.

  1. Set this equal to -2: −6k−2=−2\frac{-6}{k-2}=-2k−2−6​=−2.

  2. Solve: −6=−2(k−2)-6=-2(k-2)−6=−2(k−2), so −6=−2k+4-6=-2k+4−6=−2k+4, giving k=5k=5k=5.

Exam technique

In the exam

  1. Rearrange into y=mx+cy=mx+cy=mx+c before deciding on the gradient.

  2. For parallel lines, keep the same gradient; for perpendicular lines, flip the fraction and change the sign.

  3. If points are given, calculate m=change in ychange in xm=\frac{\text{change in }y}{\text{change in }x}m=change in xchange in y​ carefully, then state your conclusion clearly.

Self review

Check yourself

  • What gradient would a line perpendicular to y=35x−2y=\frac{3}{5}x-2y=53​x−2 have?

  • Can you rearrange 2x+3y=92x+3y=92x+3y=9 into y=mx+cy=mx+cy=mx+c and identify the gradient?

  • If two lines have gradients m=4m=4m=4 and m=−14m=-\frac{1}{4}m=−41​, what is their relationship?

Recap questions

Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.

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Parallel and Perpendicular Lines Revision Guide

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